I'm studying the comparison Hessian theorem and I not understand the following:
Let $(M, \langle\ ,\ \rangle)$ be a complete Riemannian manifold. Given $o\in M$, define $r=dist(o, \cdot)$. Then, for $r \rightarrow 0^{+}$,
\begin{equation} \operatorname{Hess}(r) = \frac1r\bigg\{ \langle\ ,\ \rangle - dr\otimes dr \bigg\} + o(1). \end{equation}
I tried use Taylor's formula for $r$. That is, if $\gamma:[0,1] \rightarrow M$ is a smooth path, such that, $\gamma(0)=o, \ \gamma'(0)=v$ and $\gamma(t)=p$, for some $t\in[0,1]$, then $$ (r\circ \gamma)(t) = (r\circ \gamma)(0) + (r\circ \gamma)'(0)t + \frac12(r\circ \gamma)''(0)t^2 + o(t^2), $$ i.e.,
$$ r(p) = \langle \nabla r, v\rangle t + \frac12 \operatorname{Hess}(r)(v,v)t^2 + o(t^2). $$
So,
$$ \frac12 \operatorname{Hess}(r)(v,v) = \frac{r}{t^2} - \frac1t \langle \nabla r, v\rangle + o(1). $$
Since $r\rightarrow 0^{+}$, I believe that
$$ \frac12 \operatorname{Hess}(r)(v,v) = \frac{1}{r} \bigg\{ 1- \langle \nabla r, v\rangle \bigg\} + o(1). $$
But, this is no concluded the equality above.
Someone has some idea? Thank you!